a: x/72=5/12

=>x/72=30/72

hay x=30

b: x+3/15=-1/3

=>x=-1/3-1/5=-8/15

c: =>-x/4=21/y=z/80=3/4

=>x=-3; y=28; z=60

kho the

\(\dfrac{x+2}{3}=\dfrac{y-5}{-4}=\dfrac{z+1}{5}\Rightarrow\dfrac{2x+4}{6}=\dfrac{3y-15}{-12}=\dfrac{z+1}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x+4}{6}=\dfrac{3y-15}{-12}=\dfrac{z+1}{5}=\dfrac{2x+4-3y+15+z+1}{6-\left(-12\right)+5}=\dfrac{\left(2x-3y+z\right)+\left(4+15+1\right)}{23}=\dfrac{72+20}{23}=\dfrac{92}{23}=4\)

\(\dfrac{x+2}{3}=4\Rightarrow x+2=12\Rightarrow x=10\\ \dfrac{y-5}{-4}=4\Rightarrow y-5=-16\Rightarrow y=-11\\ \dfrac{z+1}{5}=4\Rightarrow z+1=20\Rightarrow z=19\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x+2}{3}=\dfrac{y-5}{-4}=\dfrac{z+1}{5}=\dfrac{2x-3y+z+4+15+1}{2\cdot3-3\cdot\left(-4\right)+5}=\dfrac{92}{23}=4\)

Do đó: x=10; y=-11; z=4

\(\dfrac{x+2}{3}=\dfrac{y-5}{-4}=\dfrac{z+1}{5}\text{ và }2x-3y+z=72\)

\(\text{Áp dụng tính chất dãy tỉ số bằng nhau:}\)

\(\dfrac{x+2}{3}=\dfrac{y-5}{-4}=\dfrac{z+1}{5}=\dfrac{2\left(x+2\right)-3\left(y-5\right)+z+1}{2.3-3.\left(-4\right)+5}=\dfrac{92}{23}=4\)

\(\Rightarrow\dfrac{x+2}{3}=4\Rightarrow x+2=12\Rightarrow x=10\)

\(\dfrac{y-5}{-4}=4\Rightarrow y-5=-16\Rightarrow y=-11\)

\(\dfrac{z+1}{5}=4\Rightarrow z+1=20\Rightarrow z=19\)

a) Vì : \(\left|x+\frac{19}{5}\right|\ge0\forall x\in R\)

\(\left|y+\frac{1890}{1975}\right|\ge0\forall y\in R\)

\(\left|z-2004\right|\ge0\forall z\in R\)

\(\Rightarrow\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2004\right|\ge0\forall x,y,z\in R\)

Dấu''='' xảy ra khi và chỉ khi \(\hept{\begin{cases}x=\frac{-19}{5}\\y=\frac{-1890}{1975}\\z=2004\end{cases}}\)

b,\(\left|x+\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|\le0\)

Ta có:\(\left|x+\frac{9}{2}\right|\ge0\forall x\)

          \( \left|y+\frac{4}{3}\right|\ge0\forall y\)

          \(\left|z+\frac{7}{2}\right|\ge0\forall z\)

\(\Rightarrow\left|x+\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|\ge0\forall x,y,z\)

Mà \(\left|x+\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|\le0\)

\(\Rightarrow\left|x+\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|=0\)

\(\Rightarrow\hept{\begin{cases}\left|x+\frac{9}{2}\right|=0\\\left|y+\frac{4}{3}\right|=0\\\left|z+\frac{7}{2}\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x+\frac{9}{2}=0\\y+\frac{4}{3}=0\\z+\frac{7}{2}=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{9}{2}\\y=-\frac{4}{3}\\z=-\frac{7}{2}\end{cases}}}\)

c,\(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\)

Ta có:\(\left|x+\frac{3}{4}\right|\ge0\forall x\)

          \(\left|y-\frac{1}{5}\right|\ge0\forall y\)

          \(\left|x+y+z\right|\ge0\forall x,y,z\)

\(\Rightarrow\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x,y,z\)

Mà \(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\)

\(\Rightarrow\hept{\begin{cases}\left|x+\frac{3}{4}\right|=0\\\left|y-\frac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x+\frac{3}{4}=0\\y-\frac{1}{5}=0\\x+y+z=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{3}{4}\\y=\frac{1}{5}\\z=\frac{11}{20}\end{cases}}}\)

Chọn đáp án C

Đặt \(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c\)

\(hpt\Leftrightarrow\left\{{}\begin{matrix}a+b=\frac{1}{72}\\a+c=\frac{1}{63}\\b+c=\frac{1}{56}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=\frac{1}{72}\\a-b=\frac{1}{63}-\frac{1}{56}\\b+c=\frac{1}{56}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1}{72}-b\\2b=\frac{1}{72}+\frac{1}{504}\\c=\frac{1}{56}-b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1}{72}-b\\b=\frac{\frac{1}{72}+\frac{1}{504}}{2}=\frac{1}{126}\\c=\frac{1}{56}-b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1}{72}-\frac{1}{126}=\frac{1}{168}\\b=\frac{1}{126}\\c=\frac{1}{56}-\frac{1}{126}=\frac{5}{504}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=168\\y=126\\z=100,8\end{matrix}\right.\)